The absorption of a linearly polarized photon.

[Khrapko]

Linearly polarized light does not carry angular momentum. However, individual photons emitted, say, in the direction of the X-axis, carries a spin angular momentum \hbar in the direction + X or -X. And, when absorbing of the photon, target gets its energy h\nu and its angular momentum \hbar. Is it right?

[olgranpappy]

Linearly polarized light does not carry angular momentum. However, individual photons emitted, say, in the direction of the X-axis, carries a spin angular momentum \hbar in the direction + X or -X. And, when absorbing of the photon, target gets its energy h\nu and its angular momentum \hbar. Is it right?

no..

[DrDu]

No, the emitted photons in linearly polarized light are in a superposition state of the two helicity eigenstates.

[Andy Resnick]

Linearly polarized light does not carry angular momentum. However, individual photons emitted, say, in the direction of the X-axis, carries a spin angular momentum \hbar in the direction + X or -X. And, when absorbing of the photon, target gets its energy h\nu and its angular momentum \hbar. Is it right?

Light can have both spin as well as orbital angular momentum:

http://www.physics.gla.ac.uk/Optics/play/photonOAM/

Photons have spin momentum only, corresponding to \Delta l = \pm 1 . It's possible to have linearly polarized emission, corresponding to \Delta m= \pm 1 , but it's an unusual process that typically occurs in the presence of magnetic fields or specially designed materials.

[DrDu]

Interesting link, Andy. But it says clearly that also single photons can have an angular orbital momentum.
It is also possible to expand an electromagnetic field into vector spherical harmonics which are angular momentum eigenstates. These can have all values of total angular momentum j starting from 1. This minimal value is what is meant when speaking of the "spin" of the photon. In the narrow sense of the word, the photon doesn't have a spin, as the spin is defined as the angular momentum in the rest frame of the particle, which does not exist for light.
What is l and m ?

I don't think that one can say that the usual emission is circularly polarized. The light emitted from most lamps has no polarization at all, so it can be either viewed as a mixture of circularly polarized photons or linearly polarized photons. Most fluorescent molecules are elongated in one direction and the transition dipole moment also oscillates in that direction which results in the emission of linearly polarized photons. For preferencial circular polarization, one needs either chiral molecules or magnetic fields.

[Cthugha]

Light can have both spin as well as orbital angular momentum:


You can also have a conversion from one to the other. See for example
http://people.na.infn.it/~marrucci/oam/index.htm
and the references on that page.

[Khrapko]

Dear DrDu, it is impossible that a photon has no spin (no matter what the state of the photon).
Please consider an analogy: an isotropic emission of particles. The state is a superposition of plane wave in all direction. But an individual particle has concrete momentum.
At the same time you are right concerning vector spherical harmonics which are orbital angular momentum eigenstates. But besides this orbital angular momentum, a dipole radiation can contain spin. http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=7

[DrDu]

Dear Krapko, I just wanted to point out that speaking of the spin of a photon is strictly speaking, an abuse of nomenclature. A Photon has helicity.
If I remember correctly, the vector spherical harmonics contain both the spin and the orbital angular momentum.

[Khrapko]

DrDu,
(i) Would you please let me know the difference between spin of photon, which it does not have, in your opinion, and helicty, which it has?
(ii) It is strange that the vector spherical harmonics are the same for a photon with its spin (or helisty) 1, for an electron with its spin 1/2, and for a spinless particle, if, as you remember, the vector spherical harmonics contain both the spin and the orbital angular momentum.

[Khrapko]

no..

A photon has no spin, i.e. has spin zero, only if the photon's state is an eigenstate corresponding to the zero eigenvalue. But a linear polarization is not an eigenstate.

[DrDu]

DrDu,
(i) Would you please let me know the difference between spin of photon, which it does not have, in your opinion, and helicty, which it has?
(ii) It is strange that the vector spherical harmonics are the same for a photon with its spin (or helisty) 1, for an electron with its spin 1/2, and for a spinless particle, if, as you remember, the vector spherical harmonics contain both the spin and the orbital angular momentum.

(i)To answer your question I have to delve into group theory. According to Wigner, particles are characterized as representations of the Poincare group. The representations of massless particles like photons and massive particles like electrons differ in that the wavevector k is timelike for massive particles, i.e. it a certain frame of reference (the rest frame in that case) it has the form (mc,0,0,0)^T, while for a massless particle it can at best be brought to the form const.*(1,0,0,-1)^T. The sub-group of the Poincare group which leaves the wave-vector invariant is known as the "little group" of the wavevector. In case of massive particles it is SO(3), the rotation group which leads directly to spin, while in the case of massless particles it is E(2), the group of all translations and rotations in a plane. It has a sub-group U(1), which leads to the classification according to helicity.
(ii)Why do you think that the vector spherical harmonics, which describe e.g. particles with spin 1 also describe particles with spin 1/2 or 0?
For spin 0 we use the simpler (scalar) spherical harmonics, which are prominent as solutions of the angular dependence of the hydrogen wave functions.
For spin 1, there are three different vector spherical harmonics. For light there are only two
allowed due to transversality restriction. Due to this restriction it is no longer possible to separate the angular momentum into a sum of spin and orbital momentum.

[Khrapko]

DrDu,
(i) Well, what is a numerical value of photon's helicity?
(ii) Sorry, "the simpler (scalar) spherical harmonics, which are prominent as solutions of the angular dependence of the hydrogen wave functions", is, in reality, the electron's wave function rather than hydrogen's one; and an electron has spin 1/2. So, we are forced to introduce electron's spin besides orbital AM.
Next, consider the decay $\Lambda\to N+\pi$:
$\psi=A_s\chi_+Y_0^0+A_p(\chi_+Y_1^0/\sqrt{3}-\chi_-Y_1^1/\sqrt{2/3}$ where $\chi$ is the Pauli spinor. So, we are forced to introduce $\pi$'s spin besides the orbital AM.
I think, we must introduce photon's spin besides orbital AM. This is even more the case that the radiation patterns of orbital AM and spin are orthogonal (see Difference Between Spin and Orbital Angular Momentum http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=7)
The belief is naive that spin will be automatically introduced by the Jackson's procedure 9.6 "Spherical Wave Solution of the Scalar Wave Equation"

Sorry, I cannot write formulae but I can send attachments

[DrDu]

Yes, in case of an electron we have to take spin into account. However, we can consider wavefunctions which are a product of a spin part and the orbital angular momentum part so it is possible to consider spin and OAM separately.
The same holds true for a spin 1 particle. But for a photon, only those solutions are allowed where the field is transversal. These states are superpositions of different product states of spin and orbital angular momentum in general, so it becomes impossible to disentangle spin and orbital angular momentum. This is a well known result:
V. B. Berestetskii, E. M. Lifshitz, and L. P. Pitaevskii, Quantum Electrodynamics (Pergamon Press, Oxford, 1982).


Btw. to write latex formulas, you have to begin the formulas with "square bracket" "tex" "square bracket" and end with "square bracket" "/tex" "square bracket" instead of $ signs.

[Khrapko]

I know very well the well-known statement about the impossibility. I know you cannot disentangle spin and orbital angular momentum. But why are you sure that you have spin in the frame of Maxwell electrodynamics (quantum electrodynamics brings nothing into Maxwell electrodynamics). Naturally, spin cannot be separated from an angular momentum if the angular momentum does not contain spin. Maxwell angular momentum is moment of momentum, {\bf L}={\bf r}\times({\bf E}\times{\bf B}). But let us defer this problem.
What about the absorption of a linearly polarized photon? I repeat: A photon has no spin, i.e. has spin zero, only if the photon's state is an eigenstate corresponding to the zero eigenvalue. But a linear polarization is not an eigenstate. So, an individual linearly polarized photon does have spin or helicty.

[DrDu]

This is only true insofar as S^2 or the square of helicity is different from zero. However,
a linear polarized photon does certainly not have either positive or negative helicity as long as I am not measuring it. That's the same situation as in a double slit experiment where claiming that the particle has passed through only either of the slits leads to contradictions.

[Khrapko]

(i) Sorry, I ask a simple thing: does target get angular momentum \hbar when it absorbs one linearly polarized photon, or not? Yes or no? Can you predict?
(ii) You did not answer, what is a numerical value of photon's helicity? So, your "the square of helicity" has no sense.

[DrDu]

(i) no
(ii) The eigenvalues of helicity are plus minus hbar

Good night!

[Khrapko]

Good night!
No!
Your answers are somewhat vague
(i) Sorry, I ask a simple thing: does target get angular momentum \hbar when it absorbs one linearly polarized photon, or not? Yes or no? Can you predict?
(i) no
What “no”? You cannot predict? Or you can predict zero?
If you predict s_x=0 for our photon which is emitted in the direction of the axis X, then you do not take into account my reason that a photon has s_x=0 only if the photon's state is an eigenstate with s_x=0 eigenvalue. But a linear polarization is not an eigenstate of a photon.
Your account of the double slit experiment is needless. The target measures s_x of the photon.
Next
(ii) The eigenvalues of helicity are plus minus hbar
The eigenvalues of s_x are plus minus hbar. Does it mean that helicity = s_x? If so, then
your "the square of helicity" has no sense.

[DrDu]

(i) Sorry, I ask a simple thing: does target get angular momentum \hbar when it absorbs one linearly polarized photon, or not? Yes or no? Can you predict?
(ii) You did not answer, what is a numerical value of photon's helicity? So, your "the square of helicity" has no sense.

(i) When giving the analogy of the double slit experiment I wanted to make clear that in quantum mechanics, most questions cannot be answered as yes or no. Their answer is really undefined. (As you will jump on me anyhow, I should add that this statement only holds in standard interpretations of QM but not e.g. in nonlocal hidden variables theories which have some other strange consequences.)

Obviously I can use a beam splitter which sends photons of positive helicity upwards and those of negative helicity downwards. Then I will find in the statistical mean 50% photons of each helicity. However this destroys the information about linear polarization. If a molecule emmits a linearly polarized photon and another one absorbs it, I have no information about helicity at all.

(ii) s_x is the spin of the photon with respect to a space fixed axis and does not exist.
Helicity is the projection of angular momentum on the direction of momentum of the photon.
So spin and helicity have the same unit.

[Khrapko]

Dear DrDu, I need not your splitter. We have a linearly polarized photon, i.e. we have a photon, which passed through a linear polarizator. A target absorbs this photon. The absorption means that the target gets energy of h\nu, momentum of h\nu/c, and mass of h\nu/c^2. I ask: does the target get angular momentum of +\hbar, or of -\hbar, or zero? You cannot answer this question. So, you admit that the target maybe get zero angular momentum. Yes? Or you can say nothing?

[DrDu]

If the target was in an eigenstate of angular momentum before absorption, it will generally end up in a superposition of angular momentum states. That is, its angular momentum has not a sharp value. Nevertheless, it would be wrong to say that its angular momentum hasn't changed.

[Khrapko]

The target is not a QM object. The target is a macroscopic object. The absorption is a measurement of spin AM of the photon (in particular). The wavefunction collapse occures when absorbing. I ask, what change of AM of the target occures when absorbing: +\hbar, or -\hbar, or zero?

[Khrapko]

Obviously I can use a beam splitter which sends photons of positive helicity upwards and those of negative helicity downwards. Then I will find in the statistical mean 50% photons of each helicity. However this destroys the information about linear polarization.
Well, now replace the splitter with the target. You have to agree that 50% photons will give +\hbar and 50% photons will give -\hbar to the target!

[DrDu]

Yes, but only if you really are measuring the angular momentum. E.g. in case of a hydrogen atom, you may measure whether it got excited from 1s to 2p_+ or 2p_-. Then you will find at random one time the one and one time the other and you get an information about angular momentum. But in another experiment you may ask whether it gets excited to p_x or p_y. Depending on the linear polarization, you may find e.g. allways p_x. But then you don't have any information about angular momentum.

[Khrapko]

Yes, but only if you really are measuring the angular momentum.
Sorry, I cannot understand your thought. So I will ask in a different way.
(i) Let our macroscopic non-QM target absorbs a circularly polarized photon with p_x=h\nu/c. Do you expect a change of \hbar in J_x of the target? If yes, why? Are you really measuring the angular momentum J_x of the target? (p is momentum, J is angular momentum).
(ii) Let our macroscopic non-QM target absorbs a linearly polarized photon with p_x=h\nu/c. Do you expect a change of \hbar in J_x of the target?

[Khrapko]

I do not like alluding to authorities, but now I am forced because my reasons are not apprehended. So I cite Feynman “The Lectures” 17-11:
“Macroscopic measurements made on a beam of linearly polarized light show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amount of angular momentum - the average angular momentum is zero.”
This quotation means that each photon of linearly polarized light contributes an amount of angular momentum, i.e +\hbar or -\hbar, but not zero.

[Khrapko]

If Feynman and I are right, i.e. an individual linearly polarized photon moving in x-direction has spin angular momentum s_x=+\hbar or s_x=-\hbar, then an important question arises:
How can the angular momentum conservation law be satisfied? http://www.physicsforums.com/showthread.php?t=413566
Really,
R.Feynman (the Lectures, Quantum Mechanics, Ch. 18) considers an excited atom sitting with its angular momentum along +z-axis, i.e. s_z=+\hbar in the initial state. Then the atom emits a photon in x-direction. So, in the finish state, we have a system consists of our atom without angular momentum and a photon radiated in x-axis direction with s_x=+\hbar or s_x=-\hbar. It troubles me.

[DrDu]

If the atom is in eigenstates of s_z before and after emission, then s_x is completely undefined. Hence no problem with conservation.

[Khrapko]

s_x is completely defined up to sign because p_x is completely defined. Please be attentive,
the atom emits a photon in x-direction.
photon radiated in x-axis direction with s_x=+\hbar or s_x=-\hbar.

[Khrapko]

Dear colleagues. I submitted this question, "Is angular momentum conserved when an atom emits an individual photon?", to Am. J. Phys. on 26 Nov 2001. Jan Tobochnik rejected this submission. It seems that his reasoning is interesting, but I do not understand it completely:
“I think your question is wrong. An atom cannot emit a photon with a particular component of spin. It simply emits a photon with spin 1 and then a detector can be lined up to measure the value of the component of the spin along a certain direction. QM tells us that the spin is never perfectly alligned in any direction. There is always an uncertain amount in some other direction. If one measures the spin of the photon and that of the atom, quantum mechanics will always conspire to conserve angular momentum.”

[DrDu]

I did not doubt that the spin s_x of the photon is well defined. I said that the spin s_x of an atom in an eigenstate of s_z is undefined (as long as s^2 isn't 0).

[Khrapko]

s_z is defined (of \hbar) in the initial state. s_x is defined (of \hbar) after emitting. How can the angular momentum conservation law be satisfied: that is the question

[Khrapko]

Dear colleagues, I have a hypothesis.
We can save the angular momentum conservation law if take into account spin of the atom’s electron. Feynman ignores electron’s spin, but we must take it into account. Then the initial orbital angular momentum of the excited atom, l_z=\hbar, is emitted as an orbital angular momentum of the photon, and spin of the photon, s_x=\pm\hbar, is provided by an overturn of electron’s spin. A corollary: after emitting of a photon with defined p_x, the atom will have j_x=\pm\hbar/2.
Analogically, when an electric dipole rotating in x-y-plane emits a photon in z-direction, one electron of the dipole is being overturned. So, in other words, a rotating dipole is being magnetized in the transverse direction [1].
[1] R.I. Khrapko A rotating electric dipole radiates spin and orbital angular momentum http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=8

[Khrapko]

R.Feynman (the Lectures, Quantum Mechanics, Ch. 18) calculates the radiation pattern of an excited atom and the distribution of spin in this radiation at l = 1, ignoring the electron's spin of the excited atom. Can it really be true, these distributions do not depend on the relative orientation of the electron spin and its orbital angular momentum in the initial state of the excited atom, ie on: j = 1/2 or j = 3/2?
Does somebody know an experiment on the distributions?

[DrDu]

In general yes, but in light atoms, spin-orbit coupling is small so that the splitting of the term components with different j can be neglected.

[Khrapko]

I am not interested in the energy splitting that really can be neglected. I interested in the space distributions of spin, angular momentum, and mass in the radiation of the excited atoms with different j. Does somebody know experiments on the distributions?

[hiyok]

@ Dr Du,

I think it is incorrect when you say a photon does not have angular momentum but only spin. Actually, in some cases when the system has spherical symmetry, it is more convenient to have a photon with definite angular momentum (aside from its spin).

[DrDu]

Where did I say this?

[hiyok]

Sorry, Dr Du, I made a mistake. It is Andy Resnick who said that. Really sorry for this slip !!!

hiyok

[Khrapko]

I think it is incorrect a photon does not have angular momentum but only spin.
This sentence is strange because spin is an angular momentum.



in some cases when the system has spherical symmetry, it is more convenient to have a photon with definite angular momentum (aside from its spin).
This sentence is strange because a spherical symmetric system is described by Y_0^0=Const spherical harmonic, and has no angular momentum.
By the way, I think spherical harmonics do not describe spin

[hiyok]

@Khrapko:

(1) You may infer from the context, that I reserve angular momentum only for orbital angular momentum. While spin, although behaving as angular momentum under rotation, I use it to mean internal degrees of freedom;

(2) No. A spherically symmetric system means the communtation between rotation operator and Hamiltonian. It means the Hamiltonian and the angular momentum can be simutaneously measured with 100% certainty. It means the basis states of the system can be written as |n;l,m>, where n labels energy levels while (l,m) labels angular momentum. Y00 corresponds to only (0,0). But the system can actually take other states with non-zero angular momentum.
Spherical symmetry does not mean the wave function must be a constant.

(3)Of course, spherical harmonics can be used to describe the integer spin (but not half integers). For example, for spin 1, you use Y10,Y11 and Y1-1 as the basis states. And in this case, the angular operator can be represented by 3*3 matrice.

hiyok

[Khrapko]

A spherically symmetric system means the communtation between rotation operator and Hamiltonian.
This sentence is strange because angular momentum operators commute with Hamiltonian always independently of a system. Symmetry of a system means that an operator exists that does not change the system. A sphere is spherically symmetric, but a cylinder is not spherically symmetric. Only s-states (Y00) of an electron is spherically symmetric (if we disengage from electron spin). An electron with spin has no spherically symmetric states.


While spin, I use it to mean internal degrees of freedom;
spherical harmonics can be used to describe the integer spin
A question arises: if spherical harmonics can be used to describe spin as well as an orbital angular momentum, why do you name namely spin an internal degree of freedom?

[hiyok]

@Khrapko:
(1) As regards what is meant by symmetry and the C.R. between angular operators and Hamiltonian, we don't have to debate here. Because, this is a textbook content. You may consult any textbook in quantum mechanics.

(2)That I name it as an internal degree of freedom is largely my personal preference. Fpr example, I also name isospin as an internal degree of freedom, which can actually be described by Pauli Matrice, the same Pauli matrice that are used to describe spin half. Indeed, such matrice are even used to describe any two level system.

(3)I would say, all the above two points are textbook content.

[Khrapko]

As regards what is meant by symmetry and the C.R. between angular operators and Hamiltonian, we don't have to debate here.
We will not have the debate if you agree that “a symmetry operation on a system is an operation that does not change the system” (The Pengium Dictionary of Physics) while the angular momentum operators commute with Hamiltonian always independently of a system because of the symmetry (isotropy) of our space rather than of a system.
I also name isospin as an internal degree of freedom, which can actually be described by Pauli Matrice, the same Pauli matrice that are used to describe spin half.
Well, an internal degree of freedom (spin half) is described by Pauli Matrice and must be added to the external degree of freedom, namely, to orbital angular momentum, which is described by spherical harmonics (see #12). The question is: why they do not add the internal degree of freedom (spin one) of a photon to the external degree of freedom of the photon, namely, to orbital angular momentum, which is described by spherical harmonics? I add spin of photon to the orbital angular momentum [1].
[1] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=9

[hiyok]

(1) I would agree with you if you said 'a symmetry operation is an operation that leaves the Hamiltonian and the dynamical laws of the system invariant, but not its states.'

(2)I never said that the spin of a photon can not be added to its orbital angular momentum. I just emphasized photons posses both spin and orbital angular momentum. Surely, they can be added, since both transform in essentially the same way under spatial rotation. So, we agree here.

[Khrapko]

I would agree with you if you said 'a symmetry operation is an operation that leaves the Hamiltonian and the dynamical laws of the system invariant, but not its states.'
I have cited “The Pengium Dictionary of Physics”. Btw, you ignored my note that cylinder has no spherical symmetry, and so p-state of an electron is not spherically symmetric.
I never said that the spin of a photon can not be added to its orbital angular momentum. I just emphasized photons posses both spin and orbital angular momentum. Surely, they can be added, since both transform in essentially the same way under spatial rotation. So, we agree here.
It is excellent! Thank you very much! I struggle with the scientific community during 10 years! I introduced a spin tensor into the electrodynamics in addition to the moment of momentum. Physicists consider electrodynamics spin as a moment of a linear momentum. My papers were rejected more than 500 times (see http://khrapkori.wmsite.ru/). For example [1], Ohanian wrote: “This angular momentum is the spin of the beam” [2]
DrDu wrote (#13): “it becomes impossible to disentangle spin and orbital angular momentum. This is a well known result”
[1] http://www.physicsforums.com/showthread.php?t=418699&highlight=khrapko #13
[2] Ohanian, What is spin? Am. J. Phys. 54 (1986) p. 502

[Sir Beaver]

Just to clarify, what hiyok says is that a Hamiltonian which is spherically symmetric will commute with L^2 and Lz. This is not true for any system. Also, in physics, the statements 'Hamiltonian has this symmetry' and 'system has this symmetry' is the same thing. Finally, just because the Hamiltonian has one symmetry, does not mean that the wave function will have this symmetry. For example, the Hamiltonian in a periodic lattice is periodic with R, but not the wave function, which is a Bloch vector.

[Khrapko]

It seems to me that the angular momentum operators commute with Hamiltonian always independently of a system because of the symmetry (isotropy) of our space rather than of a system.

[Khrapko]

I never said that the spin of a photon can not be added to its orbital angular momentum. I just emphasized photons posses both spin and orbital angular momentum. Surely, they can be added, since both transform in essentially the same way under spatial rotation.

It is excellent! Thank you very much! I struggle with the scientific community during 10 years! I introduced a spin tensor into the electrodynamics
Now I have found new reasons for the classical spin tensor. I submitted a paper [1] to PRA. The paper is devoted to an irritative question: how is an angular momentum flux of electromagnetic field distributed in the space?
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

[DrDu]

Just a remark: Spin ( in systems where it is strictly defined, like for electrons but not for photons) is, strictly speaking, not angular momentum and it doesn't transform as angular momentum under spacial rotations. E. g. an electron has always spin 1/2 hbar while orbital angular momentum depends on the choice of the origin. Spin is angular momentum in the rest frame of the particle with the particles' position taken as origin.

[DrDu]

R.Feynman (the Lectures, Quantum Mechanics, Ch. 18) calculates the radiation pattern of an excited atom and the distribution of spin in this radiation at l = 1, ignoring the electron's spin of the excited atom. Can it really be true, these distributions do not depend on the relative orientation of the electron spin and its orbital angular momentum in the initial state of the excited atom, ie on: j = 1/2 or j = 3/2?
Does somebody know an experiment on the distributions?

I just want to discuss the transition starting from l=1 j=3/2 to l=0 j=1/2:
I take as initial excited state the one state with l=1 j=3/2 which is a product state |l=1, m=1 > |m_s=+1/2> (all others with same j correspond to rotated situations).
There are two possible finals states with l=0:
l=0 j=1/2: |l=0, m=0> |m_s=+1/2>
and {l=0, m=0 > |m_s=-1/2>
For light of long wavelength, transition is mainly due to the electric dipole moment d which only couples to the orbital part of the wavefunction hence the transition probability is
<l=1, m=1|\mathbf{d}| l=0, m=0> <m_s=1/2|m_s=\pm 1/2>
The last matrix element is 1 if m_s doen't change and zero else.
A single photon transition to the state with s=-1/2 would require an admixture to the initial state with j=3/2, s=-1/2 and different value of l due to spin-orbit interaction. In case of the 2p ->1s transition in hydrogen, this is completely negigible.

[Khrapko]

orbital angular momentum depends on the choice of the origin.
NO! Angular momentum of a rotating wheel does not depend of the position of an observer. Orbital angular momentum of an atom’s electron does not depend on the position of an observer as well. A fortiori, spin angular momentum does not depend on the position of an observer. Angular momentum, L, of an object depends on a displacement, dr, of the position of an observer only if the oject has a linear momentum, P: dL = P x dr.
Spin of electrons is not angular momentum and it doesn't transform as angular momentum under spacial rotations.
Only angular momentum of a wheel, which is a pseudo vector (with an outer orientation), is transformed as a vector under spatial rotations. Orbital angular momentum of an atom’s electron is a characteristic of its Q.M. state, which is a spinor. So, it is transformed not as a vector. All the more spin angular momentum of an electron, which is represented by 1/2-spinor, is not transformed as a vector. Nevertheless, orbital angular momentum of an atom’s electron is added to spin angular momentum of the electron: j = l + s.

[Khrapko]

To DrDu #51
It seems to me the matrix element of the transition from l=1, m=1 to l=0, m=0 is <l=0,m=0\mid{\bf d}\mid l=1,m=1>.
But I am interested in the difference between the transitions from m=1, m_s=1/2 and from m=1, m_s=-1/2

[DrDu]

You are right, better to say, transition probability is the absolute square of that matrix element, so the order of the bras and kets doesn't matter too much.
The transition probabilities from the j=1/2 state can be worked out in analogy. The wavefunction with j=1/2 and l=1 is of the form
a |l=1, m=1, m_s=-1/2> -b |l=1, m=0, m_s=1/2>
I don't want to determine a and b. However, they can be chosen real and positive.
The transition probability becomes
|a|^2 |<l=1, m=1|\mathbf{d}|l=0,m=0>|^2+|b|^2 |<l=1,m=0|\mathbf{d}|l=0, m=0>|^2 - 2ab Re (<l=1,m=1|\mathbf{d}|l=0,m=0><l=0,m=0|\mathbf{d}|l=1,m=0>)

There is an analogous state with j=3/2 which can be obtained by interchanging a->b and b-> -a.
As was to be expected, the transitions between states with different spin m_s become mixed up but this is not due to an explicit spin flip. In case of the state with j=3/2, even the mixing interpretation can be avoided by refering m and m_s to a rotated quantization axis.

[Khrapko]

I should like to restrict us to the Feynman’s case.
I am interested in the space distributions of spin, angular momentum, and mass in the radiation when an atom makes the transition from l=1, m_l=1, m_s=1/2 or from l=1. m_l=1, m_s=-1/2. Note these radiations have different frequencies (see #34, #36).

[DrDu]

I should like to restrict us to the Feynman’s case.
I am interested in the space distributions of spin, angular momentum, and mass in the radiation when an atom makes the transition from l=1, m_l=1, m_s=1/2 or from l=1. m_l=1, m_s=-1/2. Note these radiations have different frequencies (see #34, #36).

The transition from the first state I discussed in #51. When frequencies of the transitions are not the same for the second state, this is due to Spin-Orbit splitting. Then the second state, i.e. l=1. m_l=1, m_s=-1/2 is not an eigenstate but mixes with l=1 m_l=0 and m_s=1/2, which is the situation I discussed in #54. However, in the limit of vanishing Spin Orbit coupling, also the second state is an eigenstate. Then the transition in question does not change spin as the wavefunctions factorize into the spin and the orbital part and the transition is due to the dipole moment operator which only acts on the orbital wavefunction.

[hiyok]

Now I have found new reasons for the classical spin tensor. I submitted a paper [1] to PRA. The paper is devoted to an irritative question: how is an angular momentum flux of electromagnetic field distributed in the space?
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

Hey, Khrapko,
I have discussed your paper with my friends. And I feel that your claim (which, as I understand, is that the expression \vec{J}=\vec{r}\times(E\times B) does not include the spin, let's say, [tex]\vec{s}[\tex], which is the polarization of photons.) is not sufficiently justified for the following reason:

1. You calculated the spatial distribution of J and s and found that they are spatially separated, and then you claimed an incompatibility between J and s. However, one can express J as a sum of s and another term (which I shall call L), J=s+L. And it is likely that, the distribution of L just makes up for this spatial separation of J and s.

2. Actually, the division of J into s and L is well known[Quantum Field Theory: from operators to path integrals, Kerson Huang, John Wiley & Sons, 1998]. According to this division, s can indeed be related to the polarization of photons and yield results in accord with Feynman's doing. This means that, L is the orbital angular momentum (the moment of momentum) while J is the total one. Considering Lorentz invariance (invariant under Lorentz group), this must be so.

The reason why I believe that J has already included the s is because, the above division gives correct results.

[Khrapko]

DrDu #51 discussed the transition from l=1 j=3/2 to l=0 m_s=-1/2. It is not interesting.

DrDu #54 stated that |l=1 m=1 s=-1/2> is not an eigenstate,
that the eigenstate is |l=1 j=1/2> = a|l=1 m=1 s=-1/2> - b|l=1 m=0 s=1/2>.
It seems to be wrong because the statement gives rise to a doublet splitting of the term 2P_1/2 just as of ammonia (see Feynman’s Lectures).
Really, denote for short |l=1 m=1 s=-1/2> = F, |l=1 m=0 s=1/2> = G. Then
\hat H F=h_{11}F+h_{12}G,\qquad \hat H G=h_{21}F+h_{22}G
\hat H (aF-bG)=ah_{11}F+ah_{12}G-bh_{21}F-bh_{22}G=E(aF-bG)
ah_{11}F-EaF-bh_{21}F=0,\qquad ah_{12}G-bh_{22}G+EbG=0
(h_{11}-E)(-h_{22}+E)+h_{12}h_{21}=0
has two solutions.
However, as before I am interested in the space distributions of spin, angular momentum, and mass in the radiation when an atom makes the transition from |l=1 m=1, s=-1/2> (#34). And as before I am interested in how can the angular momentum conservation law be satisfied (#32).

Hiyok #57 agrees that the area of the circular polarization is spatially separated from the area where the moment of momentum exists, ie from the area where r\times(E\times B)\ne 0 as is stated in [1].
Hiyok #57 claims that the area where r\times(E\times B)\ne 0 can be expanded to include the circular polarization area by the division \int r\times(E\times B)dV=s+L, but it is strange because a division cannot give rise to an expansion. Unfortunately, the Hiyok’s delusion is a common delusion [2-4]. I explained the matter [5,6].
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files
[2] J. Humblet, Physica, 10, 585 (1943)
[3] J. D. Jackson, Classical Electrodynamics, Problem 7.27
[4] H. C. Ohanian, “What is spin?” Amer. J. Phys. 54, 500-505 (1986).
[5] R.I. Khrapko. True energy-momentum tensors are unique. Electrodynamics spin tensor is not zero. - http://arXiv.org/abs/physics/0102084
[6] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=9

[DrDu]

Of course you get two solutions, but the other one corresponds to j=3/2, m_j=1/2, not j=1/2. The factors a and b are the Clebsch Gordan coefficients. You can calculate them e.g. here:
http://personal.ph.surrey.ac.uk/~phs3ps/cgjava.html
a= sqrt(2/3) and b=sqrt(1/3) for the j=1/2 m_j=1/2 state.
For the j=3/2, m_j=1/2 state, you have to change a->b and b->-a. You may check this in the calculator explicitly.

Btw. your h_12 is due to spin-orbit interactions which are small. If you neglect them, the two states stay degenerate and you can work with the states F and G equivalently.

[Khrapko]

DrDu #59
Of course you are right: (J=3/2, M=1/2) = sqrt(2/3) (m=0,s=1/2) + sqrt(1/3) (m=1,s=-1/2).
Are the distributions in the radiation from (J=3/2, M=1/2)-state the superpositions of the distributions in the radiations from (m_l=0)-state and (m_l=1)-state? And the electron has no role in the distributions? Electron can be ignored when calculating?

[DrDu]

No, the radiation from the states with specified J is not just the superposition of the one from states with fixed L and M.


What do you mean with "the electron plays no role?".

[Khrapko]

What do you mean with "the electron plays no role?".
Sorry, I meant, “the electron’s spin plays no role”
As is known [Jackson], an energy distribution from an electron in the (l=1, m=1)-state is \cos^2\theta+1. Feynman’s Lectures yields this also (see [1])
The distribution from (l=1,m=0)-state is \sin^2\theta.
I think the distribution from the state sqrt(2/3) (l=1,m=0;s=1/2) + sqrt(1/3) (l=1,m=1;s=-1/2), as in #60, is the superposition of the distributions.
If no, how can we calculate the distribution?
[1] http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

[DrDu]

Well, there should be also an interference term, as you don't superpose the energy distributions but the wavefunctions. This is something like 2abFG in your nomenclature. I would have to delve into the derivation of the expressions of Feynman myself to work out these expressions.

[Khrapko]

Well, we can observe \cos^2\theta+1 power distribution from (J=3/2, M=3/2) = (l=1,m=1;s=1/2)-state.
But we cannot observe \sin^2\theta power distribution from (l=1,m=0)-state (according to Jackson, 9,9) because (l=1,m=0,s=1/2)-state is not an eigenstate?
I would have to delve into the derivation of the expressions of Feynman myself to work out these expressions.
You don’t need to delve. Feynman did not obtain the result \cos^2\theta+1. Please see (4.4) in http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

[DrDu]

You are asking complicated questions. My analysis up to now treated the field as a classical variable. Hence it is not possible to read of directly the change in the field upon emission.
To treat it quantum mechanically, one presumably has to introduce photon eigenfunctions with given l>0 and m, |l m> so that the electric and magnetic field are something like <x |E|lm> and <x|B|lm> and are products of some kind of vector spherical harmonics and Bessel functions (outgoing Hankel functions to describe a resonance). I can only sketch the reasoning: Before emission we have an excited state of the form aF +bG (both in case of the j=3/2 m_j=1/2 and the j=1/2 m_j=1/2 case) and after emission one with aF'+bG'. Here, F' and G' are obtained from F and G by replacing the excited orbital wavefunction with l and m by the corresponding photon wavefunction with l and m times the (unique) electronic ground state wavefunction with l=0 and m=0. Hence the wavefunction of the photon becomes entangled with the spin wavefunction of the electron.
However, orthogonality of the spin wavefunctions of the electron, will lead to the result that the angular distribution of the radiation emitted is simply |a|^2 (cos^2(theta)+1)+|b|^2 sin^2(theta), so there will be no interference effects in contrast to what I first assumed.

[Khrapko]

Thank you. Our discussion is helpful.
1) My conclusion is the distributions \cos^2\theta+1 and \sin^2\theta [1-3] concern the radiation of a rotating dipole and of an oscillating dipole rather than an atom’s radiation. So the problem of the separate radiation of orbital angular momentum and spin by a rotating dipole may be beyond the scope of atom physics. However,
2) Do you agree that the power distribution in the radiation from (J=3/2, M=3/2) = (l=1,m=1;s=1/2)-state of an atom is \cos^2\theta+1? Do you know if experimental evidences of the distribution exist?
[1] J. Jackson, Classical Electrodynamics, #9.9
[2] A. Corney, Atomic and Laser Spectroscopy (1979), Fig.2.6
[3] L. D. Landau, E. M. Lifshitz, The Classical Theory of Fields, #67, Problem 1

[DrDu]

The transition from the J=3/2 M_J=3/2 state to the J=1/2 M_J=1/2 state is, for small spin-orbit coupling and wavelength sufficiently larger than the atomic dimensions, a purely electronic dipolar transition. Electron's spin doesn't change.
The angular power should presumably be calculated from the wavefunction of the photon |photon>=|el, l=1, m=1> as <el,l=1,m=1|S|el, l=1,m=1> where S=c/8 pi (E x B-B x E) is the Poynting vector operator and the electronic wavefunction is somehow related to the vector spherical harmonics, the "el" referring to the polarization resulting from an electronic dipole transition.
I am not completely sure how to evaluate this expression. Presumably S has to be normal ordered. Then E and B only have matrix elements between |photon> and the vacuum |0>. Then one can introduce the vacuum state so that the matrix element become e.g.
<photon|B|0> x<0|E|photon> =-i omega <photon|B |0> x rot <0|B|photon>. I set b=<0|B|photon>. In the second step, I used Ampere-Maxwells law. I get S \propto \nabla |b|^2 -1/2 (b^*\cdot \nabla b+b \cdot \nabla b^*). From what I found, the angular dependence of b is given as \Phi_{lm}=\nabla Y_{lm}=\sqrt{3/8 \pi}\exp(i \phi)(i \hat{\theta}-\cos \theta \hat{\phi})
Putting all together, the radial component of S in deed varies like 1+cos^2(theta). However, there are also non-vanishing angular components.

[Khrapko]

the radial component of S in deed varies like 1+cos^2(theta).
Feynman obtained 1+cos^2(theta) simply (see (4.4) in [1]). Besides, Feynman calculated the distribution of spin, cos(theta) (see (4.2) in [1]). Can you obtain the distribution of spin by your method?
there are also non-vanishing angular components.
Maxwell electrodynamics gives the distribution of orbital angular momentum, sin^2(theta) (see (2.9) in [1], or (2.79) in [2]). Can you confirm this result by your method?
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files
[2] A. Corney, Atomic and Laser Spectroscopy (1979)

[DrDu]

Well, Spin density is proportional to ExA. Using E=-dA/dt and Ampere Maxwell you should be able to express this as something proportional to rot b* x rot b. Is there an easy expression for spin flux?

[Khrapko]

As I understand, your method for finding the power distribution in the atom radiation, 1+\cos^2\theta, is exactly the method used in the frame of the standard electrodynamics [1] because you use Poynting vector, E\times B, that is a component of Maxwell tensor, Ampere-Maxwell’s law, and vector spherical harmonics, though I should like to see your calculations in details. In contrast, Feynman obtained 1+\cos^2\theta simply and exclusively by quantum mechanics (see (4.4) in [2]).

there are also non-vanishing angular components.
Maxwell electrodynamics gives the distribution of orbital angular momentum, sin^2(theta) (see (2.9) in [2], or (2.79) in [3]). Can you confirm this result? I ask this for the second time.

Spin density is proportional to ExA.
Unfortunately, you share a common alogism. You use Maxwell energy-momentum tensor, not the canonical energy-momentum tensor, and in the same time you use a component of the canonical spin tensor, E\times A, which is annihilated by the Belinfante-Rosenfeld procedure.
As is well known, the canonical energy-momentum tensor T_c^{ik} is coupled with the canonical spin tensor \Upsilon_c^{jik}. So the total angular momentum density is J_c^{jik}=r^{[j}T_c^{i]k}+\Upsilon_c^{jik}. But the construction contradicts experiments [4].
As is well known, Maxwell energy-momentum tensor T^{ik} is not coupled with a spin tensor, Maxwell energy-momentum tensor is coupled with zero spin tensor. So, the construction r^{[j}T^{i]k} is an orbital angular momentum.

Do you know if experimental evidences of the distribution \cos^2\theta+1 in the radiation from (J=3/2, M=3/2) atom exist?
You don’t answer.

Is there an easy expression for spin flux?
YES, an easy expression for spin flux is well known since 2001. And what is more an experiment for a confirmation of the expression is suggested [2,7]. Unfortunately, the submissions [2,4,7] were rejected without reviewing. Gordon W.F. Drake assessed my paper [2] as “too pedagogical for the Physical Review.” Sonja Grondalski wrote only, “Your manuscript has been considered. We regret to inform you that we have concluded that it is not suitable for publication in Physical Review Letters.”

[1] J. Jackson, Classical Electrodynamics, #9.9
[2] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files
[3] A. Corney, Atomic and Laser Spectroscopy (1979)
[4] Canonical spin tensor is wrong http://khrapkori.wmsite.ru/ftpgetfile.php?id=49&module=files
[5] R.I. Khrapko. True energy-momentum tensors are unique. Electrodynamics spin tensor is not zero. - http://arXiv.org/abs/physics/0102084
[6] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=9
[7] Experiment concerning electrodynamics’ nonlocality http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=46

[DrDu]

This weekend I managed to take a quick look on Feynman's calculation.
He calculates the probability amplitude to find circularly polarized plane wave photons, i. e., helicity eigenstates.
For this kind of photons, electric field is \langle \pm, k_z|\mathbf{E}|0 \rangle=E_0 (1, \pm i, 0)^T \exp(ik_z z) (don't ask me, to which helicity the plus and minus refers) for k in z-direction. Other directions can be obtained by rotating first around the z-axis, which introduces only a phase factorn and then around the y-axis ( the rotation matrix has row vectors something like (\cos(\theta), 0, -\sin(\theta), (0, 0, 0), (\sin(\theta), 0, \cos(\theta)) ), the relevant vector becomes (up to the phase factor) (\cos(\theta), \pm i, \sin(\theta))^T .

The transition is due to the \mathbf{E}\cdot \mathbf{d} coupling in the hamiltonian in the electric dipole approximation. The transition dipole moment has to be calculated with the J=3/2, M_J=3/2 L=1 M_L=1 state and the ground state with J=1/2 M_J=1/2, L=0, M_L=0. As we saw already, spin doesn't change as the operator d only acts on the orbital part. Hence the matrix elements can easily be evaluated from the spherical harmonics. The corresponding matrix element is \langle 3/2, 3/2| \mathbf{d}|1/2, 1/2\rangle=d_0 (1, i, 0)^T (again it could be -i instead of i). The scalar product of the electric field and transition dipole moment is then proportional to 1 \pm \cos(\theta) depending on the polarization.

[Khrapko]

Thank you for the attention, but, sorry, I have understood nothing in your post. Feynman does not use E. What is the power T? How can be obtained the scalar product of the electric field and transition dipole moment? Can you recommend textbooks where the notations are used?

[DrDu]

It was not my intention to repeat Feynmans consideration word by word but to make contact with what we discussed before. I wanted to point out especially that Feynman seems to consider circular polarized plane waves, that is states of sharp helicity in his analysis.

T means "transposed", i.e. consider the column vector instead of the row vector.
The coupling to the electromagnetic field reads different in different gauges. In atomic and molecular physics it is very convenient to start from the multipolar gauge where the coupling is given in terms of a series of couplings to the different electric and magnetic multipole moments of the atom or molecule. The first and most important term is the coupling of the electric field to the electric dipole moment. This can be found in many books e.g. in the book by Craigh and Thirunamachandran, Molecular quantum electrodynamics, Dover publ.
or also here, if you know some french: http://www.phys.ens.fr/cours/college-de-france/1986-87/1986-87.pdf or in the books of Claude Cohen Tannoudji.
In the simplest cases one can argue that in the Coulomb gauge, the coupling of some electrons i to the electromagnetic field has the form \sum_i e\mathbf{A}\cdot \mathbf{p_i}. Now \mathbf{E}=-d\mathbf{A}/dt and \mathbf{p_i}=d\mathbf{r_i}/dt hence \sum_i e \mathbf{A}\cdot \mathbf{p}_i and \mathbf{E}\cdot \mathbf{d} with \mathbf{d}=e \sum_i \mathbf{r_i} coincide up to a total time derivative which can always be added to the Lagrangian.

[DrDu]

Just found on arxiv an article that might interest you:

http://arxiv.org/ftp/arxiv/papers/1010/1010.1056.pdf

[Khrapko]

I know A. M. Stewart very well since 2005. I criticized his articles [1,2,3] in 2005, but James Dimond rejected my submittion [4] to Eur. J. Ph.:
“From the confused and inaccurate Abstract right through to the end, this paper contains a large number of errors and misunderstandings. I know that similar papers by Khrapko have been rejected by a number of other journals. The lack of understanding shown by the author is beyond any easy solution; no simple re-writing of the paper can make it sensible. I strongly recommend that the paper be rejected as unsuitable for publication.”
Then I criticized Stewart’s articles in my publication [5], but Stewart ignores criticism and continues to publish his mistakes.
Unfortunately, L. Allen, M. J. Padgett [6] ignore my criticism in [5] as well.

[1] A. M. Stewart, Angular momentum of the electromagnetic field: the plane wave paradox resolved, European Journal of Physics 26, 635-641 (2005). arXiv:physics/0504082v3
[2] A. M. Stewart, Angular momentum of light, Journal of Modern Optics 52(8), 1145-1154 (2005). arXiv:physics/0504078v2
[3] A. M. Stewart, Equivalence of two mathematical forms for the bound angular momentum of the electromagnetic field, Journal of Modern Optics 52(18), 2695-2698 (2005). arXiv:physics/0602157v3
[4] R. I. Khrapko, Angular momentum of the electromagnetic field, EJP/202604/PAP/ (June 30, 2005) http://khrapkori.wmsite.ru/ftpgetfile.php?id=57&module=files
[5] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=9
[6] L. Allen, M. J. Padgett, “Response to Question #79. Does a plane wave carry spin angular momentum?” Am. J. Phys. 70, 567 (2002) http://khrapkori.wmsite.ru/ftpgetfile.php?id=53&module=files

[Khrapko]

I am sure that the disgraceful report (see #75) was written by A.M. Stewart rather than by J. Dimond, editor, because my submission [4] concerned Stewat’s paper [1]. However, since editors support anonymity of reviewers, they are responsible for reports.
I think it is the situation, which was discussed at [2]. Juan R. Gonzalez-Alvarez wrote on 9 Apr 14:08
“Biased reviewers want to maintain their names anonymous to accept the papers of their friends/colleagues and for rejecting the papers of the people working in rival theories or showing the mistakes contained in referee's work (when as author)...”
Note, arXiv publishs A.M. Stewart, but I am blacklisted [3].
[1] A. M. Stewart, Angular momentum of the electromagnetic field: the plane wave paradox resolved, European Journal of Physics 26, 635-641 (2005). arXiv:physics/0504082v3
[2] Peer-Review Under Review http://groups.google.ru/group/sci.physics.electromag/browse_thread/thread/6a35d56e15ad1b69/5711f1c2b0ce22dd
[3] http://khrapkori.wmsite.ru/files/struggle-with-arxiv-10?catoffset=10
[4] R. I. Khrapko, Angular momentum of the electromagnetic field, EJP/202604/PAP/ (June 30, 2005) http://khrapkori.wmsite.ru/ftpgetfile.php?id=57&module=files

[Khrapko]

Dear DrDu, you wrote in #54:
The wavefunction with j=1/2 and l=1 is of the form
a |l=1, m=1, m_s=-1/2> -b |l=1, m=0, m_s=1/2>

I should like to obtain the ratio a/b by the use of the way, which was started in #58, though we know a/b=sqrt{2} for (J=1/2, M=1/2), and a/b=-1/sqrt{2} for (J=3/2, M=1/2) from http://personal.ph.surrey.ac.uk/~phs3ps/cgjava.html.
Denote for short |l=1 m=1 s=-1/2> = F, |l=1 m=0 s=1/2> = G. Then
\hat H F=h_{11}F+h_{12}G,\qquad \hat H G=h_{21}F+h_{22}G
\hat H (aF-bG)=ah_{11}F+ah_{12}G-bh_{21}F-bh_{22}G=E(aF-bG)
ah_{11}F-EaF-bh_{21}F=0,\qquad ah_{12}G-bh_{22}G+EbG=0
(h_{11}-E)(-h_{22}+E)+h_{12}h_{21}=0
has two solutions.
These solutions are .
E_{1,2}=(h_{11}+h_{22})/2\pm\sqrt{(h_{11}-h_{22})^2/4+h_{12}h_{21}}.
Then
a/b=h_{21}/(h_{11}-E)=(h_{22}-E)/h_{12}.
So,
a_1/b_1=(h_{22}-h_{11})/2h_{12}-\sqrt{((h_{11}-h_{22})/2h_{12})^2+1}.
How can we obtain a/b=\sqrt{2}?
Thank you

[DrDu]

Of course you can. You should first identify the relevant hamiltonian.

[Khrapko]

Oh! The Hamiltonian depends on the energy of spin-orbital interaction. Do you mean the energy depends on the Clebsch-Gordan coefficients? It is strange because the coefficients are the expansion coefficients of total angular momentum eigenstates in an uncoupled tensor product basis; they are not connected with the energy of spin-orbital interaction.

[DrDu]

Well, when you take spin orbit interaction into account, then states with different total angular momentum will have different energies. So if you take H to equal the spin orbit part of the hamiltonian, then the clebsch gordan coefficients should result from diagonalizing the hamiltonian.

[Khrapko]

I have diagonalized the Hamiltonian. The diagonal is E_1,E_2 from #77. Respective eigenfunctions of the Hamiltonian are
\psi_1=a_1/b_1|l=1, m=1, m_s=-1/2> - |l=1, m=0, m_s=1/2> and \psi_2=a_2/b_2|l=1, m=1, m_s=-1/2> - |l=1, m=0, m_s=1/2> from #77 or from #54. The Clebsch-Gordan coefficients give a_1/b_1=\sqrt{2} and a_2/b_2=-1/\sqrt{2} where a_1/b_1 is the function of h_{ij} (see #77).
Do you assert that Clebsch-Gordan coefficients contain information about the energy of spin-orbital interaction, i.e. about h_{ij}, which are in a/b? Can Clebsch-Gordan coefficients help to identify the relevant hamiltonian?

[DrDu]

Well, in a sense they contain information about the hamiltonian, but this information you already have once you know the symmety ( in this case the rotational symmetry) of the hamiltonian.

[Khrapko]

Your post is strange. The energy of spin-orbital interaction, h_{12}, depends on e,h,c,\mu, etc., but Clebsch-Gordan coefficients are the expansion coefficients of total angular momentum eigenstates in an uncoupled tensor product basis; they are not connected with the energy of spin-orbital interaction. So, they cannot determine the energy of spin-orbital interaction, h_{12}. The equation a_1/b_1=(h_{22}-h_{11})/2h_{12}-\sqrt{((h_{11}-h_{22})/2h_{12})^2+1}=\sqrt{2} (#77) is a puzzle!

[DrDu]

Your post is strange.
Well, maybe because I don't exactly understand yours!
The hamiltonian has to interchange with J and also L^2 and S^2. This limits it's general form but of course does not determine the absolute strength of the interaction.
But apparently it fixes the ratio (h22-h11)/h12.